This Is My New Favorite Number
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- Опубліковано 2 бер 2022
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i is the imaginary (complex) unit, but what is its square root?
We'll use some complex number theory and matching coefficients to solve a system of equations to figure this one out!
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#math #brithemathguy #i
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The neat thing is, the fact that there isn't a meta-imaginary means that our number system is closed.
Yeah. It's interesting that humanity had to spent so much time to make numbers "complete". There are quaternions, though, but they don't have some properties which imaginary numbers have.
Does been able tomae undefined numbers and singularities still keep it closed?
It is still possible to define further numbers; even natural numbers are “made up.” Check out the dual numbers, a separate number system where the new unit E^2 = 0
@@Kokurorokuko imaginary numbers also lack some properties of real numbers. Inequalities dont work anymore
Sets are only closed under particular operations, so we need to specify which operations and sets we're talking about for that statement to be meaningful.
That’s pretty cool; personally, I like number 4.
lmfaooo
LMAO
Perfection
welp he's definitely not asian
SAME. Everyone always thinks it's weird, but it is important to solving math entirely. Well, every number has a role. But I like 4's role.😊😊😊
Alternative method:
Represent i in polar form
Therefore, i = cos(π/2) + isin(π/2)
By Euler's formula, e^(iθ) = cosθ + isinθ
Therefore i = e^(iπ/2)
Now √i = i^(1/2) = e^(iπ/4)
Therefore √i = cos(π/4) + isin(π/4)
√i = 1/√2 + i/√2
√i = (1+i)/√2
Yeah, that's what I thought about right when I saw the thumbnail.
@@PunmasterSTP Because it is so much easier
This is quite neat.
Also my method. Had it in three steps.
So much uglier, but works
i was actually wondering this the other day. how this recommendation was so spot on blows my mind
Literally the same for me
I was studying complex numbers and this got recommended
@@thedarkslayer9475 same
Apple is listening through our phones I swear
Same for me. I think the video was recommended based on our search histories and overall interests.
Technically we define i by having the property i^2=-1. That is, sqrt(-1) is the principal root of i^2. It's just more convenient for this video say i=sqrt(-1).
blackpenredpen found the √i using both this method and using polar coordinates.
Nice video but i is not equal to sqrt(-1) : suppose sqrt(-1) exist, so
sqrt(-1)² = sqrt(-1)×sqrt(-1)
= sqrt(-1×-1)
= sqrt(1)
= 1; wich is different of i², which is -1
So i is not a number that has a value, his only property is i²= -1, so your starting point is wrong, nice try though
@@themlaw9895 it is wrong √(-1)×√(-1)= √[(-1)×(-1)=√1=1
The property √a×√b=√ab holds if and only atleast one of the number is positive that is either a≥0 or b≥0
@@themlaw9895 The property of exponents you just used isn't true for complex numbers, but you enter that field when you talk about √-1. i is as real as real numbers, and nature works in complex numbers, not real ones.
I am excited to learn upper class math🤩, but I haven't books
Just a small thing; 0:14 Imaginary numbers are a subset of Complex Numbers, not the same.
before it was official, people used imaginary numbers to refer to any numbers that have an imaginary part
@@user-sb9ho5jz3e Official? Like what time? Cardano's time? Descartes time? Euler's time? Gauss time? Cauchy time? What is "official complex numbers"?
Another awesome video as always! I was curious about another method. Instead of using either Euler or De Moivre's formula, when you have the equation 0 + 1i = (a^2 - b^2) + 2ab*i, you could try setting a = b. If you do, then 1 = 2b^2 and so b = +/- sqrt(2)/2 = a, and sqrt(i) = +/- (sqrt(2)/2 + i*sqrt(2)/2). Knowing there are only two square roots of a complex number, you would be done. I also tried setting a = -b, and I got complex values for a and b, but plugging them into (a + bi) I got the same thing!
Cool!
Nice! I always thought of it as:
i = e^(i pi/2)
i^(1/2) = +- e^( i pi/4)
and just leave it at that, because I prefer polar form for some reason
In the complex numbers, the nth root function takes on n values. Probably important to mention that.
No. Do you know what the definition of a function is?
@@angelmendez-rivera351 Oh boy. Look what we have here. A person who seems to correct every little thing just to feel superior because they want to demonstrate that they know what they're talking about. All you really demonstrate is that you're insecure and probably terrible at that which you boast to practice with such skill. There is something called a multivalued function. This is clearly what I meant by "function." This association is hardly needed when you're talking to people who are somewhat competent. In complex analysis, the definition of function is extended. If you knew that, you wouldn't be trying to correct me. It remains a fact that the nth root is a multivalued function on C, and every nonzero complex number has n different complex nth roots.
@@PubicGore *Oh boy. Look what we have here. A person who seems to correct every little thing just to feel superior because they want to demonstrate that they know what they're talking about. All you really demonstrate is that you're insecure and probably terrible at that which you boast to practice with such skill.*
Being said by someone who is being outwardly and intentionally condescending, and putting their train of insults before any valid arguments, this is just meaningless bluster. Did you get it out of your system? Did that stroke your ego enough?
*There is something called a multivalued function. This is clearly what I meant by "function."*
I know what it is you are talking about, but your misuse of terminology was worth pointing out. "Multivalued function" is a contradiction in terms.
*In complex analysis, the definition of function is extended. If you knew that, you wouldn't be trying to correct me.*
No. I know you are wrong, and I have more than sufficient education to know that. Your ignorant self being in a state of denial does not change this fact. I know the study of Riemann sheets is important in complex analysis and projective geometry, and they are also an an important foundation for studying presheafs. It changes absolutely nothing about the fact that you are wrong. No textbook in complex analysis defines function differently than in any other context. If you actually were literate on the subject, you would never talk about "multivalued functions." Almost no one educated on the subject does.
@@PubicGore woah chill, you wrecked that guy-
Hello Brian! Your videos are captivating as usual. Really enjoyed it.
I was hoping if you could kindly make a video on tensors. It would be really helpful
you can also just know that a 45 degree, 90 degree, 45 degree triangle with hypotenuse of 1 has sides of 1/√2
this, its a beautiful way to see it
This was a problem on my first Further Maths test in AS-Levels. Pretty easy but very fun!
I was given this problem in an introductory maths module for physics and thought of a simple geometric interpretation. I figured √(i) as being half the rotation around the origin in an argand diagram from 1 to i, since multiplying by i has the effect of "rotating" a number 90 degrees counter clockwise. Therefore the coordinates for √(i) would be the points on the unit circle that were half way between 1 and i either way. This would gives two points on the unit circle, one going 45 degrees counter clockwise from 1, and another going 135 degrees clockwise from 1. You can quickly see these are both 45 degree angles from the axis, thus a^2 + a^2 = 1 --> 2a^2 = 1 --> a^2 = 1/2 --> a = +1/√(2) or -1/√(2) for the respective top right and bottom left quadrant solutions. Therefore √(i) = +1/√(2) * (1 + i) or -1/√(2) * (1 + i)
Nice 😊👍👍👍
What is the concept of rotating imaginary numbers called?
@@kaan8964 it’s really more of the general interpretation of multiplication in general. It’s a great interpretation. I recommend 3blue1brown’s video on complex numbers, he gives a wonderful way to rethink what addition and multiplication actually mean. The way to think of complex numbers falls right out of that.
@@kaan8964 complex numbers are in a form of a+bi but so is r*e^(i*theta), and theta is the rotation part. Example: i is equal to e^(i*pi/2) because it is rotated pi/2 radians from the +Real axis. -1 = e^(i*pi) because it is rotated pi radians, which is why when you multiply by negative numbers, you 'flip' to it's negative side.
Just took Complex Analysis last semester. Good vid
Also, you can think about sqrt(i) in context of rotation: -1 is turn on 180 degree and if we multiple any number by i twice, it's means we rotate by 90 degree two times and get negative number with respect to original one. So, in that case, we should to find "rotation" of real number by 45 degree. Using knowledge about trigonometry we find out length of imaginary projection and real one the same. So, the answer will be (1/sqrt(2)) * (1 + i)
I'm so happy that I managed to get the real and imaginary parts being cosπ/4 and sinπ/4 just by looking at the thumbnail, feels nice to have something make Intuit sense.
A neat way to think about this is rotation, multiplication by i gives a 90° or π/2 movement, so √i would be half that
Thanks! This is amazing!
I'm an Italian high school student and I've just watched your video. I found it amazing! If I were you I wouldn't have a clue how to solve it. Just outstanding!
This is the same as rotating 1/sqrt2 + (1/sqrt 2)i by 45 degrees to get i, which is neat this means the cube root will be 30 degrees. The third root of i is cos(30) + sin(30)i = sqrt3/2 + (1/2)i. The rule is therefore nth root of i = cos(pi/2n) + sin(pi/2n)i
Find i^i
Know: e^(iπ) = -1
Raise both sides to 1/2 th power
e^(iπ)^(1/2) = (-1)^(1/2)
e^(iπ)^(1/2) = i
Raise both sides to the i th power
e^(iπ)^(1/2)^(i) = i^i
By properties of exponents, we have:
e^(i²π1/2) = i^i
i² = -1 by definition, in the end we have:
e^(-π/2) = i^i
You can also solve for this as:
i=cos(π/2+2πn)+i*sin(π/2+2πn)
√(i)=cos(π/4+πn)+i*sin(π/4+πn) where n is an integer
Since if we have (cos(x)+i*sin(x))^(a)=cos(ax)+i*sin(ax)
Nicer to see with the "unit circle" (is this the right vocabulary here?) and multiplications of numbers on it as adding the angles
2:00 with the left equation we know a = ± b and knowing this the second equation is east to solve
Anywhere you see a 1/√2 can be replaced with the following (according to unit circle):
cos(π/4) = 1/√2
cos(7π/4) = 1/√2
sin(π/4) = 1/√2
sin(3π/4) = 1/√2
Alternatively you can get negatives:
cos(5π/4) = -1/√2
cos(3π/4) = -1/√2
sin(5π/4) = -1/√2
sin(7π/4) = -1/√2
Perhaps these could all be used as substitutes?
I tried replacing some of these with 1/√2 in the final equation at the end of the video and I graphed it but the equation looks wildly different. Thoughts?
Another super interesting thing is i raised to the i-th power
First the Euler's thing, e^(ix) = cosx + isinx. Plugging in x = π/2, e^(iπ/2) = i.
So, i^i = (e^(iπ/2))^i = e^(iπ/2 * i) = e^(-π/2)
This turns up in for example micro acoustics where the problem has both diffusion and propagation like properties.
Also, by logical reasoning, if 2ab = 1, then ab = 1/2, and therefore, there are two valid options for the values of a and b.
The only known factors of 2 are 1, the Square Root of 2, and 2 itself, as it is prime.
Therefore, if it is given that a or b != Square Root of 2, then a = 1 and b = 1/2, or vice versa.
Thus, the 2 and the 1/2 cancel out to 1, and 1^2 = 1.
Draw a unit circle. Bcoz |sqrti|=1 r=1
do positive 45 degree angle
the height is imaginary part
the base is real part
The way I found a and b was thus:
if 0 = a^2 - b^2, then a and b must be the same.
if 1 = 2ab, then ab = 1/2.
if a and b are the same and multiply to be 1/2, then each must be the square root of 1/2, AKA 1/sqrt(2).
Fun fact: All numbers, including real numbers, can be written in a+bi form
a can be anything, but b has to be 0
And what do you think about :
sqr(i)=j sqr(j)=h ?
Better insight is to use geometric polar coordinates and makes it easier to see why this works. i is a vector length 1 at 90 degrees. the square root is just same length 1 at (45 degrees) since square root of 1 for the length. Multiplying polar coordinates requires adding the angles. Therefore if I multiply vector 1 at 45 degrees by itself it results in 1 at 45+45 = 90 degrees. Therefore the square root of i is the vector 1 at 45 degrees. This vector in rectangular coordinates is now = Cos(45 deg) + iSin(45deg) = 1/sqr(2) + i/sqr(2) = 1/(sqr(2) (1+i). On the Cartesian plan it is a vector length 1 at 45 degrees. Square root of i. Likewise i^2 = 1(180 degrees) or a vector length 1 at 180 degrees since it is twice 90 degrees. You can take i to any power by just multiplying the exponent by the angle in polar coordinates and then convert to rectangular coordinates. The vector is always a length of 1. The math is easier.
Way easier using polar coordinates. i = exp(i.pi/2), thus i^(1/2) = exp(i.pi/4), which is the positive root you found. Of course it's opposite value, -exp(i.pi/4) will also be a valid solution, which yields the negative answer.
Imagine a unit circle centered on the origin of the complex plane. Start from 1 and rotate a certain angle to get to another number on that circle, for example -1, and to take its square root, take the point that you would get with half that angle. You can convert the numbers with angles by using sin for the imaginary part and cos for the real part.
Edit: You can rotate halfway both directions because square roots can be positive and negative.
The question is then which direction did you rotate. You can go the long way around the circle, or the short way (-1 had both paths be the same length) the path you took to get the destination determines whether the answer has a negative sign.
Half of 360° is 180°, but 360° is the same as 0° and half of 0° is still 0°, and 0° is not the same as 180°.
@@angeldude101 It doesn’t matter because square roots can be positive or negative.
sqrt(1) can be +1 or -1.
finished a complex analysis homework recently and stumbled on this number, this was 7 days ago? dope!
Some dude that ask: Hey what's your favorite number?
Me: √i
This is good. Although, I would have converted this into polar coordinates and solved it that way. Or... used roots of unity, where you divide the unit circle into n parts. In this case, divide the curve from (1, 0) to (0, i) into two parts both counterclockwise (which gives you the positive solution) and clockwise (which gives you the negative solution). I explain this in my roots of unity video.
RJ
It's more simple if you use the exponential form
I think an interesting method to this is realizing e^(i π/2) = i then taking the square roots of both sides
e^(i π/4) = i^1/2
From there apply eulers formula.
You miss a solution that way. If you start with e^(2nπi+iπ/2)=i, then take the square root, (to account for periodicity), then you’ll see you actually get two distinct solutions. The other is e^(5πi/4), which when you put into a+bi form gives -√2/2−i√2/2, which indeed gives i when squared.
i = e^(i(pi/2 + 2kpi)) => i^(1/2) = e^(i(pi/4 + kpi)) = cos(pi/4 + kpi) + i*sin(pi/4 + kpi) = + or - (sqrt(2)/2 + i*sqrt(2)/2), k being an integer.
:)
Love this stuff!! Complex numbers are among my favorite mathematical objects!! Thanks lots for posting!! :) :)
Glad you enjoyed it!
I think you can also work this out by first saying that the must have equal argument that add to pi/2 so it must have a argument of pi/4 or 3pi/4. And the modulus must multiply to make 1 and be equal. So it must be root(2)
You can think of this geometrically by drawing a unit circle on the complex plane.
Cool , another way to do it is by the eulers identity, e^i(a)= cos(a)+i*sin(a), if we replace a=pi/2 then cos=0 and sin=1 so e^i(pi/2)=i and we can extract the square root (elevate to 1/2) and by properties of the exponents we have that square root(i)=e^i(pi/4)
sorry for my english, i'm learning english and math with this Chanel and is cool
Nice! Thanks and thanks for watching!
sqrt(i) = a+bi has one solution, (a+bi)^2=i has 2 solutions because the square root function is defined as the positive answer to the square root of a number.
Also, there is a faster way.
0 = a^2 - b^2
a^2 = b^2
a = b
1 = 2ab
1/2 = b^2 = a^2
a & b = 1/sqrt(2)
sqrt(i) = 1/sqrt(2) + 1/sqrt(2)i
Just wanted to point that out, also it doesn’t require any powers of 4 so it’s simpler for younger audiences.
Hi there.
What a Coincidence! You and Professor Penn posted at the same time
🧐
u can also prove it using the fact that e^ix = i, where x = pi/2
2:46 Irrationalizing the denominator
You can also consider the complex definition of a "square (complex) number" - root the amplitude and half the angle.
With this definition, and the extra knowledge that i has an amplitude of 1 and an angle of π/2 (90°), we can use those two bits of information to conclude that sqrt(i) must have an amplitude of 1 and an angle of π/4 (45°)
All we have to do is use the formula for a complex point with amplitude α and angle θ: z=αcos(θ)+αsin(θ)i
If we plug in α=1 and θ=π/4 (45°), we get sqrt(i)=1/sqrt(2)+(1/sqrt(2))i
Of course, on a circle, π/2 radians is the same as -3π/2 radians, and half of that would be -3π/4. Plugging in the values gives you the negative answer.
I will say, im not sure why it was surprising to the videomaker that there was two answers? Thats how the square root "function" works for every other number (that fits on the complex plane - i dont exactly know ALL math, so maybe there is a number that has only 1 square root?)
It actually makes sense. If i^1 is a rotation by 90° or π/2 then i^(1/2) or √i is a rotation by 45° or π/4
you could do this:
i = sqrt(-1)
sqrt(i)=sqrt(sqrt(-1))
sqrt(sqrt(-1)) = (-1^1/2)^1/2
sqrt(i)=-1^1/4
this is should be an extra complex number because there is no number that can be 4th powered to a result of -1. i^4 = (i^2)^2 = -1^2 = 1
dont reałly need to do much algebra imo. in the same way i is halfway rotated to -1, √i is halfway rotated to i. so in polar coordinates, its 1∠ π/4. but if you want it in normal coords or whatever, itd be (1+i)×√2/2 just going off the unit circle
All you're doing is changing it from i = sin(90°) to sqrt(i) = +/- sin(45°). Neat.
It would be better if you also showed it on an Argand diagram and explained how multiplying numbers adds their arguments so the two solutions are halfway around to i in each direction
i always love your videos. could you plz make a dedicated video on phi or golden ratio.?? how it came? how to prove golden ratio? i would be very happy if you make a video about this. love from bangladesh
Another way to look at this; consider imaginary numbers as scaling and rotating numbers on the complex plane. i is at the 90 degree mark. Half a rotation from 1 to i would be a 45 degree rotation, which is at the same number as shown in the video.
Moivres's formula
Z^n = |Z|^n * cis(n*ø)
Say that Z = i, then Z^(1/2) = i^(1/2) = k
Z also equals to a+bi, so a=0 and b=1
Angle is 90° and |Z|=±1
k = ±cis(90*(1/2) = ±cis(45) = ±[(2^(1/2))(1+i)]/2
I remember back in high school, we had to put the square root of I in the form a + bi. Still makes my brain hurt to this day
what we must be aware is that
u(θ) = i^(θ ∙ 2/π)
so, for any given angle θ
u(θ) returns a complex unit in its angle.
so i^2 = -1 and i^(1/2) = (1+i)√2/2
i³ = -i and i ⁻¹ = -i too
using the properties of square roots, and the fact that i=sqrt(-1), you can assume that sqrt(i) would just be the fourth root of -1, which is also an imaginary number, therefore the most simplified way of writing it is actually just how you wrote it, which is as sqrt(i)
He makes a video of all the things I think about
I was curious about it but I immediately remembered e^(i\pi/4)
Just u can use de moivre, nothing complex, the z^(1/n) for every naturaln n and complex z (maybe except zero, but 0^x=0 anyway) can be calculated.
0:22 correction: all complex* numbers can be written in this form
2:00 that is a massively overcomplicated way of solving it. Just say a^2 = b^2, since 1 = 2ab you know they have the same sign so you know a=b, thus you can just say that a=sqrt(0.5) and you've solved it. Sqrt(i) = sqrt(0.5) + i*sqrt(0.5)
Ikr 1=2a²
1/2 = a²
a = b= 1/√2
When squaring both sides of an equation, be careful not to introduce extraneous roots.
It is the 45 degree rotation and at 45 + 180 degree rotation with absolute value 1
you could say 1/√2 = √2/2 = cos π/4
so the number is a unit with angle of π/4 or 5π/4
the same as ∓e^(πi/4)
Its cool that you used solid math to completely prove something that someone with a small amount of trig skill could have just intuitively deduced.
If -1 is a 180 degree rotation on the complex plane and the square root of -1 is i (which is a 90 degree rotation on the complex plane), it stands to reason that the square root of i would be (1/root2) + (1/root2)i. Which is a 45 degree rotation on the complex plane.
I wonder if that means that the 4th root of i is a 22.5 degree rotation? And so on.
Proofs can be so elegant sometimes.
Edit: YEP! The pattern holds!
4th root of i = cos(pi/8) + sin(pi/8)i
8th root of i = cos(pi/16) + sin(pi/16)i
and so on.
So you could generalize this to:
Nth-Root of -1 = cos(pi/n) + sin(pi/n)i
Neat!
That's basically De Moivre's theorem
@@WaterMeetsLava Oh! Good to know! I'll look it up!
√2i is superior
Because since a+b= √(a²+2ab+b²)
If a=1 and b=i
It's 1+i= √(1+2i-1)
So 1+i= √2i
i is defined such that i² = -1
Technically sqrt(-1) is not allowed since square roots only take non-negative numbers
the whole point of i is to allow rooting negative numbers.
@@833Rowan There is a difference between rooting negative numbers (which is still not allowed) and finding solutions for equations such as x² = -1
The square root always maps from non-negative to non-negative real numbers.
There’s an easier way to do this. I solved it in a few seconds in my head using polar/exponential form.
sqrt(i) = exp(i*pi*(1/4+N)). So obviously there are two answers, corresponding to pi/4 and pi*5/4
3:22 you aren't supposed to keep root in the denominator
That's why my teacher doesn't like square roots out of non negative real numbers: since it gives you two different values, it cannot be used as a function
There's a more intuitive and geometric way of thinking about it: if a and b are complex numbers on the Argand-Gauss plane, then c = a * b is the vector whose norm is the product of the norms of a and b, and whose argument is the sum of the arguments of a and b. In this case, a = b = sqrt(i) so if r is the the argument o sqrt(i), then r^2 = 1 and hence r = 1. This means that sqrt(i) must lie on the unit circle. So all we need to do now is find the vector on the unit circle such that its argument is half the argument of i(which is pi/2). That gives us c = cos(pi/4) + sin(pi/4)i, the "positive" answer. If course if c^2 = i, then so does (-c)^2 and we're done.
Hi Cabutchei. Thanks for sharing your thoughts. If interested in math competitions, please consider
ua-cam.com/video/Kw7CcLcCUe4/v-deo.html
and other videos in the Olympiad playlist. Hope you enjoy 😊
sqrt(x) is a function of x so it can only have one result, not two.
Yes, and actually using the square root symbol is abuse of notation when dealing with complex numbers. A better way of putting the problem is "Solve the quadratic equation x^2=i" and that equation has....TWO (complex) solutions of the form a+bi.
In mathematics, a square root of a number x is a number y such that y2 = x
So, by definition, you have two solutions
@@mathisnotforthefaintofheart How is it an abuse of notation? You can always just take the root with the smallest angle, even if you have the 5th root of i or any root of i, or any root of any complex number.
@@wiggles7976 Then that needs to be clearly mentioned in the question. Very often it isn't.
2:00 “Pick your favorite way of solving this”. Doesn’t it just include lesser steps to conclude from a^2-b^2=0 that a=+/-b?
You could also try 1 divided by i
1/i=(1*i)/(i*i)=i/i²=i/-1=-i
Very useful video, I had previously thought that sqrt(i)=-i.
No, because (-i)²=-1 like i²=-1
@@Ostup_Burtik Which I realized after watching the video.
in class i suggested adding "imaginary numbers" to the natraul, integer etc. chart and some student started screaming "A GAJILLION ZILLION!!!" and i did not sleep well that night...
It is not entirely true... the principle root, or square root means the one with the biggest real part. That is, sqrt(4) is always 2 and not -2 even though (-2)^2=4. Using that logic, sqrt(i) can only be the positive one
Bro. Just show it as it’s principle root i^1/2 and change it to e^ln(i^1/2) then with properties of exponents we get e^ln(I)/2, solve ln(I) (which is i(Pi/)) then we have a e^i(pi/2/2) high is them converted into e^i(Pi/4), and using Euler’s formula we have cos(Pi/4)+isin(Pi/4). The cool part is cos(Pi/4) and sin(Pi/4) are the same (the square root of 0.5) so we just have sqrt(0.5)+isqrt(0.5). Sounds hard but avoid s the rush of algebra sea.
1/sqrt(2) is also sqrt(0.5) by the way.
Shouldn't the only correct answer be when a and b are both positive? I feel like the reason you get +/- is because you started by squaring both sides, which added another solution
You will realize it is not extraneous if you plug it into the original equation though!
The sqrt(-i) is 1-i and -1+i As can be seen there is no only positive or only negative rule to be followed
yo I know im prob wrong with my logic since mathemeticians didn't use something even close to as simple as this but what if you think of it as the square root of I = the square root of the square root of -1. Then, put it = x. Square to cancel out the big square root, the square root of -1 = x^2. Then, square that to get -1 = x^4. Then, take the fourth root of both to get i + or - 1 = x. So, the square root of i = i + or - 1. If this somehow does work then ig it doesn't matter cuz its not a real number like your solution
I'm kind of proud I did it myself
How did you pass from 0=(1/4b^2)-b^2 to 0=1-4b^4 ?
This mightve just been my algebra teacher being picky, but arent you not supposed to leave roots in the denominator? I mean, i understand why 1/sqrt(2) looks better than sqrt(2)/2, but i believe the latter is easier to work with
place equation
i^2 = -1
Do square on both side again
i^4 = 1
now do root of 4 on both side and
i = 1?
u can`t do this
Here's a quick solution: Multiply by 2/2 inside square root. √i = √(i x 2/2) = √(2i) / √2 = (1 +i)/√2. Done.
Please increase the volume... It's too low..
Complex analysis is my favorite math branch.
One of mine too!
Fun fact: The German term is the misleading “Funktionentheorie”, “theory of functions”.
To me it always was too rigid - every continuously differentiable function is infinitely continuously differentiable, where’s the fun in that? Look at all those real function that are differentiable but not even continuously differentiable.
@@magicmulder wait, don't get me wrong I'm just a high schooler but I thought only continuous functions can be differentiable throughout their domain?
i prefer the polar form
I am a little confused at the end. If a = +/- 1/sqrt(2) and b = +/- 1/sqrt(2), there should be 4 solutions, not just 2. The formula +/[1/sqrt(2) + i/sqrt(2)] is too limited, as it incorrectly restrict the 2 terms to have the same sign.
By the way, the solution could be found alternatively by applying the Euler formula. In that case, I expect we would explicitly find that the solution set must include an infinite series of solutions that exhibit multiples of 2n pi
If you check all the solutions you propose, you’ll see two of them don’t work. Those are only potential solutions. If you look at the second equation, 1=2ab, it’s clear that a and b must have the same sign, otherwise 2ab would be negative, which it’s not since 2ab=1.
Also, yeah you can use Euler’s formula, but all the solutions you get from all the different exponents correspond to the same two points. You don’t actually get any new solutions, just different representations of the same solutions.
Loved it. But it is an odd universe that lets imaginary numbers have a real effect on the physics calculations.
Square Root of NEGATIVE (Square Root of -1).
Sqrt(i) = sqrt(0.5)+isqrt(0.5)
The imaginary unit, just like other mathematical constants like Euler's number, are not written like common variables.
If you watch this video and feel it difficult to understand, think about it from the opposite. When "something" becomes squared, it becomes " i ". When squared, it was 1 or -1 to be 1. Once squared, it was i or -i that would be -1.
Now, if you think of the real number as a complex number with the imaginary part as 0, that is, 1 =1 +0*i, -1 =-1 +0*i, 1^2=(1+0*i)^2=1^2+(0*i)^2+2*1*(0*i)=1^2=1, (-1)^2=(-1+0*i)^2=(-1)^2+(0*i)^2+2*(-1)*(0*i)=(-1)^2=1.
And, if you think of the imaginary number as a complex number with the real part as 0, that is, i =0 +1*i, -i =0+1*(-i), i^2=(0+1*i)^2=0^2+(1*i)^2+2*(0)*(1*i)=(1*i)^2=(i)^2=-1, (-i)^2={0+1*(-i)}^2=0^2+{1*(-i)}^2+2*(0)*{1*(-i)}={1*(-i)}^2=(-i)^2=-1:
Likewise, thought of i=(a+bi)^2 in the video method.